# Min (or Max) $L_1$ norm of Hermitian matrices with given eigenvalues

Suppose the real numbers $(\lambda_1, \ldots, \lambda_n)$ are given.

All Hermitian matrices that have eigenvalues $\lambda_i$, $i=1,\ldots,n$, can be represented in the following way:

$H = \sum_{i=1}^n \lambda_i \hat{e}_i\hat{e}_i^\dagger$

for some complete set of orthonormal vectors $\{\hat{e}_i\}$.

The $L_1$ norm of a matrix $H_{i,j}$ is the absolute sum of all of its entries $\lVert{H}\rVert_1 =\sum_{i,j}|H_{i,j}|$

My question is the following:

Amongst the set of Hermitian matrices with given eigenvalues $(\lambda_1, \ldots, \lambda_n)$, do we know which one minimizes or maximizes the $L_1$ norm?

Another way to phrase this problem is the following:

What unitary achieves $\max_{U \in U(n) } UDU^\dagger$ and $\min_{U \in U(n) } UDU^\dagger$ where $D$ is the diagonal matrix $\mathrm{diag}(\lambda_1, \ldots, \lambda_n)$ and $U(n)$ is the set of $n$ dimensional unitaries?

So far, it seems that for positive matrices, with eigenvalues $(p_1, \ldots, p_n)$, the $L_1$ norm is minimized by the diagonal matrix $\mathrm{diag}(p_1, \ldots, p_n)$. This is because the trace is invariant under unitary operations, so the $L_1$ norm is minimized when all the off diagonal terms are zero. I do not know how to generalize this result. I also do not know when the norm is maximized.

Thanks in advance for any help.

• The title says "identical eigenvalues", which I have not found in the text. In what sense are them "identical"? Maybe you mean "given"? Please, add also your own approach, how did you try to solve it. – A.Γ. Nov 3 '15 at 7:47
• I was referring to the minimum and maximum of the $L_1$ norm within the set of hermitian matrices with the same set of eigenvalues. I have changed it so hopefully it is clearer now. I have also added what I already know about the problem. I am hoping to have a deeper insight, if any are available. Many Thanks. – Bobby Nov 3 '15 at 12:48